In Riemann-Stieltjes Integration why do we take $\alpha$ to be monotonically increasing?

Question

In Riemann Stieltjes integral for the function $f$ with respect to $\alpha$ why do we always take the function $\alpha$ to be monotonically increasing? Cant we take the function $\alpha$ to be monotonically decreasing?

Answer 1

Usually integration is positive functional, which means it maps positive functions to positive numbers. But it can be made negative if you want to. Note however that assuming monotonicity of function is strong regularity assumption. For such function one sided limits exist at every point - so they can only have jump discontinuities. This regularity is crucial for setting up the integral, as far as I recall.

Answer 2

Generally in $\displaystyle \int_a^b f\,d\alpha$ one takes $\alpha$ to be a function of bounded variation. That is defined as follows. Consider $$ \sup_P \sum_i |\alpha(x_{i+1}) - \alpha(x_i)| $$ where the supremum is over all partitions $P$ of the interval $[a,b]$, such a partition being a sequence $a=x_0<x_1<x_2<\cdots<x_n=b$. This supremum is the total variation of $\alpha$ on the interval $[a,b]$. To say that $\alpha$ has bounded variation on $[a,b]$ means that the total variation is finite.

One can show that if $\alpha$ is of bounded variation then $\alpha=\beta-\gamma$ for some functions $\beta$ and $\gamma$ that are nondecreasing. In that case $$ \int f\,d\alpha = \int f\,d\beta-\int f\,d\gamma. $$ So just defining it for non-decreasing functions is nearly enough.

Answer 3

The need of taking $\alpha$ monotonically increasing stems from the fact that otherwise we would get negative differences:

$$\Delta \alpha_i :=\alpha(x_{i+1})-\alpha(x_{i}). $$

This could result in $L(P,f,\alpha) \geq U(P,f,\alpha)$, where $L(P,f,\alpha)$ is the lower Riemann sum and $U(P,f,\alpha)$ is the upper Riemann sum.